In the book called "Evaluating Feynman Integrals" by Smirnov, the following formula is given for partial fraction decomposition (p.35, eq. (3.10)):

$$\frac{1}{(x+x_1)^{a_1}(x+x_2)^{a_2}}=\sum_{i=0}^{a_1-1} \binom{a_2 - 1+i}{a_2} \frac{(-1)^i}{(x_2-x_1)^{a_2+i}(x+x_1)^{a_1-i}}+\sum_{i=0}^{a_2-1} \binom{a_1 - 1+i}{a_1} \frac{(-1)^i}{(x_2-x_1)^{a_1+i}(x+x_1)^{a_2-i}} \tag{1}$$

The author makes the following comment below (p.35):

If one of the indices, e.g. $a_2$ is non-positive, a similar decomposition is performed by expanding $(x+x_2)^{-a_2}$ in powers of $x +x_1$.

I am facing such a case, but I cannot really understand what is meant by the expansion. I could do a Taylor expansion I guess:

$$\begin{align}\frac{1}{(x+x_1)^{a_1}(x+x_2)^{a_2}} &=\frac{1}{(x+x_1)^{a_1}} \sum_{k=0}^{\infty} \frac{(x+x_1)^k }{k!}\frac{\partial^k}{\partial x^k} (x+x_2)^{-a_2} \rvert_{x=-x_1} \\ &= \sum_{k=0}^{\infty} \frac{(x+x_1)^{k-a_1} }{k!}\frac{\partial^k}{\partial x^k}(x+x_2)^{-a_2} \rvert_{x=-x_1} \tag{2}\end{align}$$

But now the power of $x+x_1$ is negative for $k<a_1$ and positive otherwise, which complicates the situation in the case where an integral must be performed afterwards. The infinite sum is also not so great. Moreover, the author says to do a "similar decomposition", which seems unneeded here. How can I do a decomposition in the case of $a_2<0$?


if $a_2 < 0$, letting $n = -a_2 > 0$, what you are expanding is $$\frac {(x + x_2)^n}{(x+x_1)^{a_1}}$$

The idea the author is expressing is that if we let $u = x + x_1$, then $x + x_2 = u + b$ where $b = x_2 - x_1$. And we can rewrite the expression as $$\frac {(u + b)^n}{u^{a_1}} = \sum_{i=0}^n{n\choose i}u^{i-a_1}b^{n-i} \\=(x_2-x_1)^{-a_2}\sum_{i=0}^{-a_2}{n\choose i}\frac 1{(x - x_1)^{a_1 - i}(x_2-x_1)^i}$$

To answer your question in the comment below, where $x, x_1, x_2$ are vectors in a real inner product space, where $(x+x_i)^{a_i}$ is replaced by $\|x + x_i\|^{a_i}$.

Again, let $u = x+x_1, b = x_2 - x_1$, and now let $n = -a_2/2, m = a_1/2$. We can now rewrite the expression as $$\frac{\langle u + b, u+b\rangle^n}{\langle u,u\rangle^m}= \frac {\left(\langle u,u\rangle + 2\langle u,b\rangle + \langle b,b\rangle\right)^n}{\langle u,u\rangle^m}$$ This expands to $$\sum_{k=0}^n\sum_{i=0}^k\binom n k\binom k i 2^{k-i}\langle u,u\rangle^{i-m}\langle u,b\rangle^{k-i}\langle b,b\rangle^{n-k} = \sum_{i+j+k = n} \frac{n!}{i!j!k!}2^j\langle u,u\rangle^{i-m}\langle u,b\rangle^j\langle b,b\rangle^k$$

However here I do not see a way to get all dependency on $u$ (and therefore on the variable $x$) in the denominator. I'm not sure what use this would be for your purposes.

  • $\begingroup$ Thank you, this is very helpful! I have one more question, which I hope is trivial: if $x$, $x_1$, $x_2$ are 4-vectors (in Euclidean space) and $a_i = 2 n_i$ with $n_i \in \mathbb{N}$ such that one can write e.g. $(x+x_1)^{a_1} = [(x+x_1)\cdot(x+x_1)]^{n_1}$, would that affect the formula that you proposed? $\endgroup$ – Jxx Oct 22 '19 at 10:54
  • $\begingroup$ Unfortunately, that changes everything. I've added some development to the idea in the post, but didn't get very far. $\endgroup$ – Paul Sinclair Oct 23 '19 at 1:15
  • $\begingroup$ Oh that's too bad, thank you very much though! Since this is in an integral, maybe I should first try to perform the angle integral and then only scalars would remain. Thanks a lot for the help anyway, hopefully this is helpful to others as well! $\endgroup$ – Jxx Oct 23 '19 at 12:06

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